### --oOoOoOoOoOoOo--'s blog

By --oOoOoOoOoOoOo--, history, 5 weeks ago,

1335E2 - Three Blocks Palindrome (hard version)

In this problem, the solution outlined by the editorial is $O(n*200*200)$, here $N$ <= $200000$.

How is the solution able to pass with that time complexity on a constraint like this?

• +8

 » 5 weeks ago, # |   +2 You pick a number (for first and third block) and iterate on its positions. Irregardless of the 'number' of numbers, there are only N positions in total to pick, thus the complexity is O(N * 200). Of course you have to implement it in an appropriate way. Similar to how, when you DFS/BFS, you iterate on the children of a node, and you do it for every node! It seems O(N^2) but there are only N distinct nodes to visit in total.